We want to point out that in [1]} there is a typo in Eq.(5.28): the second term in the sum should be proportional to \(m\)  and not to \(m^2\) .
To perform the computation we made use of Eq.(REF ), and of the following 2D Fourier integrals which can be found in [1]}:
\(\int \frac{d^2q}{(2\pi )^2}e^{-i\mathbf {q}\cdot \mathbf {b}}\log \mathbf {q}^2 = -\frac{1}{\pi b^2} \ ,\qquad \int \frac{d^2q}{(2\pi )^2}e^{-i\mathbf {q}\cdot \mathbf {b}}\log ^2 \mathbf {q}^2 = \frac{4}{\pi b^2}\log \frac{2}{b} \ .\) 
